Class 6 Exam  >  Class 6 Notes  >  Mathematics (Maths) Class 6  >  RS Aggarwal Solutions: Whole Numbers (Exercise 3E)

RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6 PDF Download

Q.1. Divide and check your answer by the corresponding multiplication in each of the following :
(i) 1936/16
(ii) 19881/47
(iii) 257796/341
(iv) 612846/582
(v) 34419/149
(vi) 39039/1001
Ans. (i) By actual division, we have
RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6
Check : 121 × 16 = 1936.
(ii) By actual division, we have
RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6
Check :
RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6
(iii) By actual division, we have
RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6
Check :

RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6

(iv) By actual division, we have
RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6

Check :

RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6
(v) By actual division, we have
RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6
Check :
RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6

(vi) By actual division, we have
RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6
Check :
RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6

Q.2. Divide and find out the quotient and remainder. Check your answer :

(i) 6971/47
(ii) 4178/35
(iii) 36195/153
(iv) 93575/400
(v) 23025/1000
(vi) 16135/875
Ans. (i) By actual division, we have

RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6

∴ Quotient = 148, Remainder = 15
Check : Dividend = Divisor × Quotient + Remainder
= 47 × 148 + 15
= 6956 + 15
= 6971
(ii) By actual division, we have
RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6
∴ Quotient = 119, Remainder = 13
Check : Dividend = Divisor × Quotient + Remainder
= 35 × 119 + 13
= 4165 + 13
= 4178
(iii) By actual division, we have
RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6
∴ Quotient = 236, Remainder = 87
Check : Dividend = Divisor × Quotient + Remainder
= 153 × 236 + 87

= 36108 × 87
= 36195
(iv) By actual division, we have
RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6
∴ Quotient = 233, Remainder = 375
Check : Dividend = Divisor × Quotient + Remainder

= 400 × 233 + 375
= 93200 × 375
= 93575
(v) By actual division, we have
RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6
∴ Quotient = 23, Remainder = 25
Check : Dividend = Divisor × Quotient + Remainder
= 1000 × 23 + 25
= 23000 × 25
= 23025
(vi) By actual division, we have
RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6
∴ Quotient = 18, Remainder = 385
Check : Dividend = Divisor × Quotient + Remainder
= 875 × 18 + 385

= 15750 × 385
= 16135

Q.3. Find the value of :
(i) 65007/1
(ii) 0/879
(iii) 981 + 5720/10
(iv) 1507 – 625/25
(v) 32277/(648 – 39)
(vi) 1573/1573 – 1573/1573
Ans.
(i) We know that any number (non-zero)
divided by 1 gives the number itself
∴ 65007/1 = 65007
(ii) We know that 0 divided by any natural
number gives 0
∴ 0/879 = 0
(iii) 981 + 5720/10 = 981 + (5720/10)
= 981 + 572
= 1553
(iv) 1507 – 625 25 = 1507 – (625 25)
= 1507 – 25
= 1482
(v) 32277/(648 – 39)
= 32277/609
RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6
∴ 32277/(648 – 39) = 53
(vi) 1573/1573 – 1573/1573
= (1573/1573) – (1573/1573)
= 1 – 1 = 0

Q.4. Find a whole number n such that n/n = n.

Ans. We have n/n = n
let n = 1, 1/1 = 1

1 = 1
which is true
Hence 1 is the required whole number.

Q.5. The product of two numbers is 504347. If one of the numbers is 317, find the

other.
Ans. Product of two numbers = 504347
One number = 317
∴ Other number = 504347/317
RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6
RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6
∴ Other number = 1591

Q.6. On dividing 59761 by a certain number, the quotient is 189 and the remainder is 37. Find the divisor.
Ans.
Here Dividend = 59761, Quotient = 189
Remainder = 37
We know that Dividend = Divisor × Quotient + Remainder
∴ 59761 = Divisor × 189 + 37
59761 – 37 = Divisor × 189
59724 = Divisor × 189
Divisor × 189 = 59724
Divisor = 59724/189
RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6
∴ Divisor = 59724/189 = 316

Q.7. On dividing 55390 by 299, the remainder is 75. Find the quotient, using the division algorithm.

Ans.
Here dividend = 55390,

Divisor = 299 and Remainder = 75
By division algorithm, we have
Dividend = Quotient × Divisor + Remainder
∴ 55390 = Quotient × 299 + 75
55390 – 75 = Quotient × 299
55315 = Quotient × 299
Quotient × 299 = 55315
∴ Quotient = 55315/299
RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6
∴ Required quotient = 185

Q.8. What least number must be subtracted from 13601 to get a number exactly divisible by 87 ?
Ans.
On dividing 13601 by 87, we have

RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6
It is clear that if we subtract 29 from 13601, the resulting number will be exactly divisible by 87.
∴ The required least number = 29.

Q.9. What least number must be added to 1056 to get a number exactly divisible by 23 ?
Ans.
Here dividend = 1056, Divisor = 23
By actual divison, we have
RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6
It is clear that if we add 2 to 21, it will become 23 which is divisible by 23.
∴ Required least number = 2.

Q.10. Find the largest 4-digit number divisible by 16.
Ans.
Greatest 4-digit number = 9999
RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6
On, dividing by 16, we get remainder as 15
∴ The required largest 4-digit number
= 9999 – 15 = 9984

Q.11. Divide the largest 5-digit number by 653. Check your answer by the division algorithm.
Ans. Largest number of 5-digits = 99999
On dividing 99999 by 653, we have
RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6
∴ Quotient = 153, Remainder = 90

Check : By division algorithm
Dividend = Divisor × Quotient + Remainder
= 653 × 153 + 90
= 99909 + 90
= 99999

Q.12. Find the least 6-digit number exactly divisible by 83.
Ans. The least 6-digit number = 100000
On dividing 100000 by 83, we have
RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6
RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6
It is clear that if we add 15 to 68, it will become 83 which is divisible by 83.
∴ Required least 6-digit number
= 100000 + 15 = 100015

Q.13. 1 dozen bananas cost Rs. 29. How many dozens can be purchased for Rs. 1392 ?
Ans. Cost of 1 dozen bananas = Rs. 29
∴ Bananas can be purchase in Rs. 1392
RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6
= 1392/29 = 48 dozens

Q.14. 19625 trees have been equally planted in 157 rows. Find the number of trees in each row.
Ans. Total number of trees = 19625

Total number of rows = 157
∴ Number of trees in each row
= 19625/157

RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6
∴ Number of trees in each row = 125

Q.15. The population of a town is 517530. If one out of every 15 is reported to be

literate, find how many literate persons are there in the town.
Ans. Total population of the town = 517530 Since there is one educated person out

of 15.
∴ Total number of educated persons in the town = 517530/15
RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6
∴ Total number of educated persons in the town = 34502.

Q.16. The cost price of 23 colour television sets is Rs. 570055. Determine the cost price of each TV set if each costs the same.
Ans.
Cost of 23 colour TV sets = Rs. 570055
Cost of 1 colour TV set
RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6
∴ Cost of 1 colour TV set
= Rs. 570055/23 = Rs. 24785

The document RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6 is a part of the Class 6 Course Mathematics (Maths) Class 6.
All you need of Class 6 at this link: Class 6
94 videos|347 docs|54 tests

Top Courses for Class 6

94 videos|347 docs|54 tests
Download as PDF
Explore Courses for Class 6 exam

Top Courses for Class 6

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

MCQs

,

Important questions

,

Exam

,

Viva Questions

,

ppt

,

shortcuts and tricks

,

study material

,

RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6

,

Free

,

past year papers

,

Summary

,

RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6

,

Semester Notes

,

practice quizzes

,

video lectures

,

Extra Questions

,

mock tests for examination

,

Previous Year Questions with Solutions

,

pdf

,

Sample Paper

,

RS Aggarwal Solutions: Whole Numbers (Exercise 3E) | Mathematics (Maths) Class 6

,

Objective type Questions

;